Studentt test2-120315062933-phpapp02

Student’s t-test
“Student” was W. S. Gossett. He
published his test anonymously as
‘Student’ because he was working for
the brewer’s Guinness and had to keep
the fact they were suing statistics a
secret

Student’s t-test
The test is used to compare samples
from two different batches.
It is usually used with small (<30) samples
that are normally distributed.

There are two types of t-test:
Matched pairs and independent pairs
If there is some link between the data then use the
matched pairs test (eg a ‘before’ and ‘after’)
If there is no link between the data then use the
independent pairs test

DFO 8 12 11 6 9 11 7 8 10 9
DFO+
ninhydrin
10 15 12 6 13 14 9 9 15 12
Question
In an investigation to determine the effectiveness of sequencing of fingerprints
10 prints are taken enhanced with DFO and then with ninhydrin.
The points of detail at each stage are recorded.
Is there a difference at the 95% confidence level?

t-test for matched pairs
1. Set up the null and alternative hypothesis:

H0 there is no difference in the number of minutae when using ninhydrin
HA there are more minutae observed after the enhancement of ninhydrin.

H0 there is no difference in the number of minutae when using ninhydrin
HA there are more minutae observed after the enhancement of ninhydrin.
This is a one-tail test.
We are testing at the 95% or 5% (0.05) level

1. Set up the null and alternative hypothesis
2. Calculate the difference between the pairs in the sample

DFO 8 12 11 6 9 11 7 8 10 9
DFO+ninhydrin 10 15 12 6 13 14 9 9 15 12
Difference (D)

DFO 8 12 11 6 9 11 7 8 10 9 Total
DFO+ninhydrin 10 15 12 6 13 14 9 9 15 12
Difference (D) 2 3 1 0 4 3 2 1 5 3 24

3. Calculate the mean of the differences

= 2.4

4. Calculate the standard deviation of the difference

= 1.51

5. Calculate the standard error

= = 0.478

6. Calculate the value of t

= = 5.0

7. Calculate the number of degrees of freedom (DoF) and find the critical value

DoF = No of pairs of data – 1

DoF = No of pairs of data – 1
10-1 = 9
= n – 1

7. Calculate the number of degrees of freedom (DoF) and
8. find the critical value

7. Calculate the number of degrees of freedom (DoF) and 8. find the critical value

9. Determine if there is a difference or not

t > tcritical (5.0 > 1.833)

t > tcritical (5.0 > 1.833)
So, the null hypothesis is rejected and the alternative
hypothesis is accepted.
The ninhydrin does make a positive difference

t-test independent samples
If there is no before and after relationship between the samples then
the independent samples test is used.

Example
Some brown dog hairs were found on the clothing of a victim at a crime scene
involving a dog.
The five of the hairs were measured: 46, 57, 54, 51, 38 μm.
A suspect is the owner of a dog with similar brown hairs. A sample of the hairs
has been taken and their widths measured: 31, 35, 50, 35, 36 μm.
Is it possible that the hairs found on the victim were left by the suspect’s dog?
Test at the %5 level.
[From D. Lucy Introduction to Statistics for Forensic Scientists Chichester: Wiley, 2005 p. 44.]

1. Calculate the mean and standard deviation for the data sets

Dog A Dog B
46 31
57 35
54 50
51 35
38 36
Total
Mean
Standard
deviation

Dog A Dog B
46 31
57 35
54 50
51 35
38 36
Total 246 187
Mean 49.2 37.4
Standard
deviation
7.463 7.301

2. Calculate the magnitude of the difference between the two means
49.2 – 37.4 = 11.8
.

3. Calculate the standard error in the difference
.

2. Calculate the magnitude of the difference between the two means.│
.

.
=
= 4.669 ≈ 4.67 (3 sf)

.

.
t = difference between the means ÷ standard error in the difference
4. Calculate the value of t:

.
t = difference between the means ÷ standard error in the difference
4.669 = 2.527
≈ 2.53 (3 sig fig)
11.8

.
5. Calculate the degrees of freedom

.
5. Calculate the degrees of freedom = n1 + n2 - 2

.
5. Calculate the degrees of freedom = n1 + n2 - 2
5 + 5 - 2 = 8

.
6. Find the critical value for the particular significance you are working to
from the table

.
and find the critical value from the table

.
from the table
At the 0.05 level tcrit = 2.306

.
If t < critical value then there is no significant difference between the two sets of data
If t > critical value then there is a significant difference between the two sets of data

.
If t < critical value then there is no significant difference between the two sets of data
If t > critical value then there is a significant difference between the two sets of data
So, at 0.05 level there is a significant difference
between the two data sets.
So it could not come from the same dog.

.
Group A
(leading questions)
Group B
(control)
2 3
3 5
4 0
4 3
2 4
4 0
3

.
Group A
(leading questions)
Group B
(control)
2 3
3 5
4 0
4 3
2 4
4 0
3
Mean = 3.143
SD= 0.8997
Mean = 2.50
SD = 2.07

.
Group A
(leading questions)
Group B
(control)
2 3
3 5
4 0
4 3
2 4
4 0
3
Mean = 3.143
SD= 0.8997
Mean = 2.50
SD = 2.07
Diff of means
= 0.64
SE of diff = 0.861
DoF = 11
t = 0.7433

.
DoF = 11
t = 0.7433

.
DoF = 11
t = 0.7433
t < tcritical at 5% level, 2-tail
There is no significant difference between the data test

Studentt test2-120315062933-phpapp02

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